Question

The resistances in Figure (a) and Figure (b) are all 6.1 Ω, and the batteries are ideal 16 V batteries. (a) When switch S in Figure (a) below is closed, what is the change in the electric potential V₁ across resistor 1, or does V₁ remain the same (enter "0")? (b) When switch S in Figure (b) below is closed, what is the change in V₁ across resistor 1, or does V₁ remain the same (enter "0")?

S R R R2 WW R R2 (a) (b)

Answer 1

a )) Rz fi Given both R, & R₂ is 6.1 r v - 16 volt. when switch s is open, current passes through only R 1 I = v Ri 16 I- 6.1 2.62 3 Amp And potential drop through R1 (Vpp) = 16 volt initial Now when you close the switch Total resistance Rret - 6.1 x 6.1 bilt bil rnet 3.os u so Total current I 5.245 Amp fret

the since both fl and fe is same current will divide equally > Il 2.623 (Xxx) final I, RL 2.623 X 6.1 (Ve) final Yet) final 16 Volt so change Ri change in electric potential across (VRD) Cat) initial final Cowa RI (AV) 16-16 (ov) = 0 volt R3 off P2 16V when switch s is open, 16 1:31 1 Amp I = RitR3 PL potential across Ik Ves); initial

= 1.31 1 X 6.1 CRs) initial (Yes initial 8 Volt Now when switch close the you Net Resistance 6.1 x 6.1 +6.1 R = net 12.2 Rnet = g.iswi I = v Rnet 16 1.7486 Amp I = g.is Now potential across RL (VRS) final 16-IR3 = 16 - 1.7486 x 6-1 10.666 5.3.3 4 (Ved) final = 16 (VP. 2) final resistor change of voltage Cours) = Vps) findet across Net) iritical 5:33 4 - 8 (BVR 1) = ove) -2.666 volt